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STRUCTURE OF Te-120, Te-122, Te-123, Te-124, Te-125, Te-126, Te-128, Te-130
By Prof.Lefteris Kaliambos (Natural Philosopher in New Energy) ( July 2014) Historically the discovery of the assumed uncharged neutron (1932) along with the invalid relativity (EXPERIMENTS REJECT RELATIVITY) led to the abandonment of the well-established electromagnetic laws, in favour of various contradicting nuclear theories, which could not lead to the nuclear structure. Under this physics crisis and using the charged UP and DOWN quarks , discovered by Gell-Mann and Zweig, I published my paper “Nuclear structure is governed by the fundamental laws of electromagnetism ”(2003), which led to my discovery of the new structure of protons and neutrons given by proton = + 5d + 4u = 288 quarks = mass of 1836.15 electrons neutron = + 4u + 8d = 288 quarks = mass of 1838.68 electrons The paper was also presented at a nuclear conference held at NCSR "Demokritos" (2002). In this photo I present the electromagnetic laws governing the nuclear structure, but a student of Einstein (Dr Th. Kalogeropoulos ) criticised my discovery of nuclear force and structure by believing that the nuclear structure is due to the invalid relativity. In fact, here one can see the 9 charged quarks in proton and the 12 ones in neutron able to give the charge distributions in nucleons for revealing the strong electromagnetic force for the nuclear binding in the correct nuclear structure by applying the laws of electromagnetism. You can see my papers of nuclear structure in my FUNDAMENTAL PHYSICS CONCEPTS . Note that according to my discovery of the LAW OF ENERGY AND MASS the mass defect in the nuclear structure is due to the photon mass of the emitting dipolic photon presented at the international conference "Frontiers of fundamental physics" (1993) organised by the natural philosophers M. Barone and F. Selleri , who gave me an award including a disc of the atomic philosopher Democritus. Nevertheless today many physicist continue to apply not the well-established laws but the various fallacious nuclear structure models which lead to complications. Nuclear structure of Tellurium with 32 blank positions Naturally occurring tellurium on Earth consists of eight stable isotopes. Two of these have been found to be radioactive: Te-128 and Te-130 undergo double beta decay. The longest-lived artificial radioisotope is Te-121 with a half-life of nearly 19 days. Comparing the Tellurium of 52 protons (even number ) with the Sn of 50 protons (even number) one sees that the structure of tellurium has the same high symmetry as that of Sn ( See my STRUURE OF Sn-112...Sn-124 ). Here the additional n52p52 with the p49n49 of Sn makes the third alpha particle existing in front of the central parallelepiped, while the n50p50 and p51n51make the fourth alpha particle existing behind the central parallelepiped. Using the three cases of the top view one must add also the symmetrical blank positions . So one concludes that the number N of blank positions is given by The two squares give 8n The first and sixth plane give 4(n) The second and fifth plane give 4{n} +8n The third and fourth h. plane give 8(n) That is N = 8n + 4(n) + 4{n} + 8n + 8(n) = 32 extra neutrons of opposite spins Or N = 4{n} + 16n + 12(n) = 32 extra neutrons of opposite spins STRUCTURE OF Te-120, Te-122, Te-124, Te-126, Te-128, Te-130 WITH S = 0 Since the 52 protons and 52 neutrons give S=0 one concludes that the S=0 of the above stable nuclides is due to the number of extra neutrons with opposite spins . For example the Te-120 with 16 extra neutrons has 4{n} + 12n with opposite spins, while the Te-130 with 26 extra neutrons has 4{n} + 16n + 6(n) with opposite spins. STRUCTURE OF Te-123 AND Te-125 WITH S =+1/2 Here the spin S=+1/2 of the above stable nuclides Is due to the fact that they have a number N of extra neutrons of negative spins, while the extra neutrons of positive spins have a number N+1. Thus the Te-123 with 19 extra neutrons has 10 extra neutrons with positive spins and 9 extra neutrons with negative spins, while the Te-125 with 21 extra neutrons has 11 extra neutrons of positive spins and 10 extra neutrons of negative spins. DIAGRAM OF TELLURIUM FORMING 32 BLANK POSITIONS Here you see the p47n47 along with the p48n48, which make two symmetrical alpha particles of opposite spins . But you cannot see the additional p49n49 the n39p39 of the third alpha particle and the n50p50 and the p51n51 of the fourth alpha particle. Also the p41, n41, p42, n42, p43, n43, p44, and n44 which form the central parallelepiped of opposite spins are not shown. Also the 8 deuterons of opposite spins from p13n13 to p20n20 and the 4 deuterons from p33n33 to p36 n36 are not shown. Moreover the extra neutrons 4(n) of the first and the sixth plane are not shown, while the extra neutrons 4n existing over the p31 and p32, and under the p21 and p22 with the extra 4n existing near the p23, p24, p29 and p30, are shown. ' n40.......p40.......n ' ' n........p38..........n38 H. Square with n' ' n31………p12.........n12.......p32' ' p31........n11.........p11…… n32 Sixth H. plane' ' n........ p29.........n10.........p10…… n30' ' n29………..p9..........n9 …….p30..........n Fifth H. plane' ' p47.......n27.........p8..........n8.........p28......... n48' ' n45.......p27.........n7..........p7........n28..........p46 Fourth H. plane' ' n47......p25.........n6.........p6..........n26...........p48' ' p45......n25……….p5..........n5……….p26.........n46 Third H. plane' ' n23………p4........n4………….p24...........n' ' n........p23……..n3………p3………..n24 Second H.plane' ' p21.........n2………p2............n22' ' n21........p1........n1.........p22 First H. plane' ' n.......p37......n37 ' ' n39.....p39..........n ' H. Square with n TOP VIEW OF THE FIRST HORIZONTAL PLANE IN WHICH ALL NUCLEONS ARE SHOWN ' HERE THE FIRST EXTRA NEUTRON (n ) MAKES THE TWO RADIAL BONDS WITH p22 AND p33 WHILE THE SECOND ONE MAKES THE TWO RADIAL BONDS WITH p21 AND p34 ' (n)........p34....... n34 ' p21....... n2........ p2....... n22 ' ' n21.........p1. .......n1.......p22 ' ' n33.......p33..... (n)' TOP VIEW OF THE SECOND HORIZONTAL PLANE Here the n near the p14 fills the blank position formed by p51 and p14. While the {n} near the p14 fills the blank position formed by p14, p24 and p44. The 2n near p24 and p23 fill the blank positions formed by p24 and p48 as well as by p23 and p45. Moreover the blank position of {n} near p23 is formed by p23, p13 and p41. Finally the blank position of n near the p13 is formed byp13 and p49. That is we have 2{n} +4n and the same situation occurs at the fifth horizontal plane. ' n' ' n14.......p14........{n}' ' n23.......p4........n4.........p24......n' n.......p23........n3........p3.........n24 ' {n}.......p13......n13' ' n ' ' ' ' ' TOP VIEW OF THE THIRD HORIZONTAL PLANE WITH POSITIVE SPINS ' '''HERE YOU SEE THE p41, n43, n42 AND p44 WHICH MAKE THE SQUARE OF THE CENTRAL PARALLELEPIPED. YOU SEE ALSO THE p45 AND n47 ALONG WITH THE n46 AND p48 WHICH REPRESENT THE DEUTERONS OF THE TWO ALPHA PARTICLES. AT THE SAME PLANE ARE SHOWN ALSO THE DEUTERONS LIKE n16p16 AND p15n15 ALONG WITH THE ADDITIONAL p49, n52, n50, AND p51 WITH 4(n) USED FOR CONSTRUCTING THE NUCLEI FROM P=53 TO p =60. HERE YOU CAN SEE ALSO THE 4(p) USED FOR CONSTRUCTING THE SAME NUCLEI. ' (p).........n50.......p51......(n) ' ' (p)........n42........p16.......n16......p44...........(n)' ' n47........p25........n6.........p6........n26.........p48''' ' p45........n25........p5.........n5........p26........ n46' ' (n).........p41.......n15.......p15.......n43..........(p)' ' (n)........p49.......n52.......(p)' Category:Fundamental physics concepts